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3 years ago

What is the purpose of the wire coil in an electromagnet?

Physics

1 answer:

ziro4ka [17]3 years ago

3 0

It is to conduct electricity in the magnet so it has an electric field.

Please BRAINLIEST!

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A motorcycle has a speed of 30 m / s. After braking, it decelerates with constant deceleration A = -3.0 m / s ^ 2

marin [14]

$a = \displaystyle\frac{v-u}{t}$

$-3 = \displaystyle\frac{v - 30}{3}$

$v -30 = -9$

$v = 21 m/s$

4 0

3 years ago

Read 2 more answers

The Gulf Stream off the east coast of the United States can flow at a rapid 3.9 m/s to the north. A ship in this current has a c

Alex Ar [27]

Answer:

72.54 degree west of south

Explanation:

flow = 3.9 m/s north

speed = 11 m/s

to find out

point due west from the current position

solution

we know here water is flowing north and ship must go south at an equal rate so that the velocities cancel and the ship just goes west

so it become like triangle with 3.3 point down and the hypotenuse is 11

so by triangle

hypotenuse ×cos(angle) = adjacent side

11 ×cos(angle) = 3.3

cos(angle) = 0.3

angle = 72.54 degree west of south

3 0

3 years ago

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State of motion is described by

Allisa [31]

Answer:

positions of motion is the answer as moving is a motion

3 0

1 year ago

A Carnot heat engine has an efficiency of 0.400. If it operates between a deep lake with a constant temperature of 298.0k and a

tatuchka [14]

Answer:

496.7 K

Explanation:

The efficiency of a Carnot engine is given by the equation:

$\eta = 1 - \frac{T_H}{T_L}$

where:

$T_H$ is the temperature of the hot reservoir

$T_C$ is the temperature of the cold reservoir

For the engine in the problem, we know that

$\eta = 0.400$ is the efficiency

$T_C = 298.0 K$ is the temperature of the cold reservoir

Solving for $T_H$ , we find:

$\frac{T_C}{T_H}=1-\eta\\T_H = \frac{T_C}{1-\eta} =\frac{298.0}{1-0.400}=496.7 K$

6 0

3 years ago

Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of

max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

$Q =\frac{KA(\delta T)}{L}$

$Q =\frac{25.87*1*20}{1}$

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0

3 years ago

Read 2 more answers